For turbulent flow of a relatively low-viscosity liquid in steel pipes (Re > 2100), an empirical heuristic for the optimum inside diameter is:\n\nGiven symbols:\nQ = volumetric flow rate (ft^3/s)\nρ = fluid density (lb/ft^3)\nDi = optimum inside pipe diameter (inches)\nμ = dynamic viscosity (centipoise, when used)

Difficulty: Medium

Correct Answer: Di,opt = 3.9 * Q^0.45 * ρ^0.13

Explanation:


Introduction / Context:
Early-stage pipe sizing often uses heuristics that relate a near-economic optimum diameter to flow rate and fluid properties. These correlations balance pumping power (friction losses) against installed cost (material and fabrication) and are useful for screening before detailed head-loss economics are performed.


Given Data / Assumptions:

  • Turbulent flow regime in steel pipe.
  • “Low” viscosity relative to water-like services in preliminary design.
  • U.S. customary units as stated in the options.


Concept / Approach:
Optimum diameter scales positively with flow rate; higher density slightly favors a larger diameter due to higher mass flow at a given Q. Strong exponents on viscosity are uncommon in turbulent-flow heuristics because friction factor is only weakly dependent on μ in fully rough regimes. Thus, correlations with extreme viscosity exponents are not realistic for the intended screen.


Step-by-Step Solution:

Screen options for plausible exponents: moderate Q exponent (≈0.36–0.50) and weak ρ dependence are typical.Eliminate formulas with very large μ exponents (physically implausible for turbulent flow).Select Di,opt = 3.9 * Q^0.45 * ρ^0.13 as the reasonable heuristic.


Verification / Alternative check:
Back-of-the-envelope checks with common flows yield diameters in practical ranges (e.g., 2–12 inches) using the selected correlation, consistent with project heuristics.


Why Other Options Are Wrong:

  • (b), (c), (d): Excessive viscosity exponents imply unrealistic sensitivity for turbulent flow, contradicting friction factor behavior.


Common Pitfalls:
Treating heuristics as code; final sizing should use detailed pressure-drop and lifecycle cost analysis for the selected service.


Final Answer:
Di,opt = 3.9 * Q^0.45 * ρ^0.13

More Questions from Process Equipment and Plant Design

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion